At the moment of release each plane has the same speed of 186 m/s, and each tank is at the same height of 3.11 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15.0° above the horizontal and the other is flying at an angle of 15.0° below the horizontal. Find the (a) magnitude and (b) direction of the velocity with which the fuel tank hits the ground if it is from plane A. Find the (c) magnitude and (d) direction of the velocity with which the fuel tank hits the ground if it is from plane B. In each part, give the direction as a positive angle with respect to the horizontal.
vx is the same for both (186cos15)
vy is either positive or negative 186sin15. Use 2gy = vf^2-vi^2 to find the two vy's
Then pythag and tan-1
To find the magnitude and direction of the velocity with which the fuel tank hits the ground for both planes, we need to break down the initial velocity of each plane into its horizontal and vertical components.
Given:
Speed of each plane (v) = 186 m/s
Height of each plane (h) = 3.11 km = 3110 m
Angle above horizontal (θ) = 15.0°
Angle below horizontal (ϕ) = -15.0°
(a) Finding the magnitude and direction of velocity for plane A:
For plane A, the angle is above the horizontal, so we use trigonometry to find its horizontal and vertical components.
Horizontal component of velocity (v₀x) = v * cos(θ)
Vertical component of velocity (v₀y) = v * sin(θ)
Substituting the given values:
v₀x = 186 m/s * cos(15.0°) ≈ 178.43 m/s
v₀y = 186 m/s * sin(15.0°) ≈ 48.29 m/s
To find the magnitude of the velocity, we use the Pythagorean theorem:
Magnitude of velocity (v₀) = √(v₀x² + v₀y²)
Substituting the values:
v₀ = √(178.43² + 48.29²) ≈ 188.36 m/s
To find the direction, we use trigonometry:
Direction of velocity (θ') = atan(v₀y / v₀x)
Substituting the values:
θ' = atan(48.29 m/s / 178.43 m/s) ≈ 15.55°
Therefore, for plane A, the magnitude of the velocity with which the fuel tank hits the ground is approximately 188.36 m/s, and the direction is approximately 15.55° above the horizontal.
(b) Finding the magnitude and direction of velocity for plane B:
For plane B, the angle is below the horizontal, so we use trigonometry to find its horizontal and vertical components in a similar manner as above.
Horizontal component of velocity for plane B (v₀x) = v * cos(ϕ)
Vertical component of velocity for plane B (v₀y) = -v * sin(ϕ) (negative sign as it is below the horizontal)
Substituting the given values:
v₀x = 186 m/s * cos(-15.0°) ≈ 178.43 m/s
v₀y = -186 m/s * sin(-15.0°) ≈ -48.29 m/s
Magnitude of velocity for plane B (v₀) = √(v₀x² + v₀y²)
v₀ = √(178.43² + (-48.29)²) ≈ 188.36 m/s
Direction of velocity for plane B (ϕ') = atan(v₀y / v₀x)
ϕ' = atan(-48.29 m/s / 178.43 m/s) ≈ -15.55°
Therefore, for plane B, the magnitude of the velocity with which the fuel tank hits the ground is approximately 188.36 m/s, and the direction is approximately 15.55° below the horizontal.
To summarize:
(a) For plane A, the magnitude is approximately 188.36 m/s, and the direction is approximately 15.55° above the horizontal.
(b) For plane B, the magnitude is approximately 188.36 m/s, and the direction is approximately 15.55° below the horizontal.